1,170 Actions

How to see $\Gamma(\mathscr{O}_S,\operatorname{Proj} S)$ as a ring?I wanted to find a natural structure morphism $A\to \Gamma(\mathscr{O}_S,\operatorname{Proj} S)$, so perhaps I can show that the maps $A\to \Gamma(\mathscr{O}_S,D(f))$ are compatible thus lifts to a morphism $A\to \Gamma(\mathscr{O}_S,\operatorname{Proj} S)$. In this way I could create the map while avoiding the specific structure of $\Gamma(\mathscr{O}_S,\operatorname{Proj} S)$. And that gives me a morphism $\operatorname{Proj}S\to \operatorname{Spec} A$. Is it correct?

Functions on reduced schemes are determined by their values at each point.@KReiser Haha I see now. In the affine case the image of $a$ lies in the maximal ideal $\mathfrak{m}_p$ means exactly $a\in p$, and the intersection of all primes is the nilradical. Therefore $a = 0$. Now a scheme is covered by affine opens, by the unique gluing property of sheaves, $a = 0\in\mathscr{O}_X(X)$ since its image in every affine open is $0$.